Question

Se um ponto se move em uma reta coordenada com aceleração a(t) e as suas
condições iniciais dadas, determine S(t).

Answer 1

4.1

$a(t)=2-6t;\ v(0)=-5;\ s(0)=4$

$a=\dfrac{\mathrm{d}v}{\mathrm{d}t}\Longrightarrow \mathrm{d}v=a\mathrm{d}t\Longrightarrow \int{\mathrm{d}v}=\int{a(t)\mathrm{d}t}$

$\Longrightarrow v(t)=\int{(2-6t)\mathrm{d}t}\Longrightarrow v(t)=2t-3t^2+C_0$

$v(0)=-5\Longrightarrow C_0=-5\ \therefore\ \boxed{v(t)=-3t^2+2t-5}$

$v=\dfrac{\mathrm{d}s}{\mathrm{d}t}\Longrightarrow \mathrm{d}s=v\mathrm{d}t\Longrightarrow \int{\mathrm{d}s}=\int{v(t)\mathrm{d}t}$

$\Longrightarrow s(t)=\int{(-3t^2+2t-5)\mathrm{d}t}\Longrightarrow s(t)=-t^3+t^2-5t+C_1$

$s(0)=4\Longrightarrow C_1=4\ \therefore\ \boxed{s(t)=-t^3+t^2-5t+4}$

4.2

$a(t)=3t^2;\ v(0)=20;\ s(0)=5$

$a=\dfrac{\mathrm{d}v}{\mathrm{d}t}\Longrightarrow\mathrm{d}v=a\mathrm{d}t \Longrightarrow\int{\mathrm{d}v}=\int{a(t)\mathrm{d}t}$

$\Longrightarrow v(t)=\int{(3t^2)\mathrm{d}t}\Longrightarrow v(t)=t^3+C_0$

$v(0)=20\Longrightarrow C_0=20\ \therefore\ \boxed{v(t)=t^3+20}$

$v=\dfrac{\mathrm{d}s}{\mathrm{d}t}\Longrightarrow \mathrm{d}s=v\mathrm{d}t\Longrightarrow \int{\mathrm{d}s}=\int{v(t)\mathrm{d}t}$

$\Longrightarrow s(t)=\int{(t^3+20)\mathrm{d}t}\Longrightarrow s(t)=\dfrac{t^4}{4}+20t+C_1$

$s(0)=5\Longrightarrow C_1=5\ \therefore\ \boxed{s(t)=\dfrac{t^4}{4}+20t+5}$